Difference between revisions of "2008 AMC 8 Problems/Problem 24"

(Solution)
 
(2 intermediate revisions by 2 users not shown)
Line 7: Line 7:
 
== Video Solution ==
 
== Video Solution ==
 
https://www.youtube.com/watch?v=ZqPFm9cU0MY  ~David
 
https://www.youtube.com/watch?v=ZqPFm9cU0MY  ~David
 
== Video Solution by Soo ==
 
https://youtu.be/IUh28kMLLVE - Soo, DRMS, NM
 
  
 
==Solution==
 
==Solution==

Latest revision as of 17:34, 6 October 2024

Problem

Ten tiles numbered $1$ through $10$ are turned face down. One tile is turned up at random, and a die is rolled. What is the probability that the product of the numbers on the tile and the die will be a square?

$\textbf{(A)}\ \frac{1}{10}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{11}{60}\qquad\textbf{(D)}\ \frac{1}{5}\qquad\textbf{(E)}\ \frac{7}{30}$


Video Solution

https://www.youtube.com/watch?v=ZqPFm9cU0MY ~David

Solution

The numbers can at most multiply to be $60$. The squares less than $60$ are $1,4,9,16,25,36,$ and $49$. The possible pairs are $(1,1),(1,4),(2,2),(4,1),(3,3),(9,1),(4,4),(8,2),(5,5),(6,6),$ and $(9,4)$. There are $11$ choices and $60$ possibilities giving a probability of $\boxed{\textbf{(C)}\ \frac{11}{60}}$.

See Also

2008 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png